An entry level confusion about spacetime. I understand that a 4-vector describes a point or event in spacetime. But I've also read (Bertschinger, 1999) that re spacetime "we are discussing tangent vectors that lie in the tangent space of the manifold at each point". If a point/event is described by a single 4-vector, what are all these tangent vectors that lie on the same point? Do they have different co-ordinates to the 'point/event' 4-vector? Could I also ask how a 4-vector 'contains' any sense of direction (I'm thinking here of a vector having direction and magnitude)?

2 Answers
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Your confusion comes from the difference between special and general relativity. In special relativity, the space-time manifold is assumed to carry the structure of 4-dimensional Minkowski space, which has the nice property that it is canonically identified with its own tangent space at the origin (since it is a vector space). So in special relativity you can speak of a space-time event as a 4-vector, and you can also speak of global Lorentz transformations (by doing the local Lorentz transformation in the tangent space and propagating the transformation to the whole space-time using the canonical identification).

In general relativity, however, the space-time is allowed to be an arbitrary Lorentzian manifold (what we do is break the global Lorentz symmetry of Minkowski space and require it to only hold infinitesimally, i.e. on the tangent space), and you don't have a canonical way of identifying the entire space-time with the tangent space of a fixed point. Therefore you cannot speak of a space-time event (now just a point in your space-time manifold) as a 4-vector!

Edit Let me try to make the difference between an affine and non-affine space more apparent.

Let us start by considering a two dimensional manifold. In fact, we'll just compare the usual flat plane and the sphere.

On the plane, we can fix an arbitrary point and call it the origin. Starting from this origin, pick a direction, and call it the $x$ direction. Consider the following two paths.

First you go 1 unit in the $x$ direction. Turn left for 90 degrees. travel another unit.

First you face the $x$ direction. Now turn left for 90 degrees, travel one unit. Turn right now for 90 degrees and travel another unit.

These two operations take you to the same place.

On the sphere, however, you can again fix an arbitrary point and call it the "origin", and a direction which we call "$x$". Now consider

First you go a quarter the way around the sphere in the $x$ direction. Turn left for 90 degrees, travel another quarter of the circumference.

First you face $x$ direction, turn left for 90 degrees, travel a quarter of the circumference. Now turn right and travel a quarter of the circumference.

(Try to trace this out on a tennis ball or on an orange.) You don't end up at the same place!

Now, going back to the plane, if instead you start by going "face $x$. Turn left for 45 degrees. Travel for $\sqrt{2}$ units." You will again end up at the same place as the previous two tries.

But on the sphere, if you start by "face $x$. Turn left for 45 degrees. Travel for $\sqrt{2}$ times a quarter circumference." You will end up somewhere else entirely again compared to the previous two tries.

What does this mean? On a flat plane, from the above demonstration, after fixing an origin and a direction $x$, we can uniquely describe every point on the plane as "certain units in the $x$ direction, and then certain units to the left of that". And it doesn't matter at all in which orders you zigzag to the destination. This identifies the plane with a vector space. That is, you can say that "the displacement from point $p$ to point $q$ is the vector $\vec{v}$, and the displacement from point $q$ to point $r$ is the vector $\vec{w}$, so the displacement from the point $p$ to point $r$ is the sum of vectors $\vec{v} + \vec{w} = \vec{w} + \vec{v}$". So after fixing a point as the origin, we can describe all other points as a vector displacement relative to this origin, and add and subtract the vectors as appropriate.

On the surface of the sphere (a curved manifold), however, the above example shows that the intuition we learned from studying classical mechanics about using vector addition for the "displacement vector" cannot hold for the positions of points. So you shouldn't think about space-time events in general relativity (which is a point on some possibly curved manifold), as vectors relative to a fixed origin. (Because with vectors you would be tempted to add them and so forth.)

I think I understand you as saying that as spacetime in SR has an origin we can therefore talk about 4-vectors as events with reference to that origin. But there is no single origin in GR spacetime and thus we can't talk about global 4-vectors. Still not sure what an event is in GR but thanks for your answer.
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Peter4075Jun 17 '11 at 20:06

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@user4075: it's not about origin at all. The Minkowski space is an affine space and doesn't have an origin either. You have to pick one arbitrarily (and that's why the identification of the Minkowski space with the tangent space is not canonical, contrary to what @Willie writes). You can in the same way pick some random point on arbitrary manifold and say it is origin. The important point here is that Minkowski is an affine space and so we can form differences of the space-time points and these are vectors. But there is no way to form these differences on general manifold.
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MarekJun 18 '11 at 16:06

@Marek: good catch. I meant of course relative to a fixed origin (I was implicitly treating Minkowski space as a vector space).
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Willie WongJun 18 '11 at 16:15

@Willie: sure, I know what you meant and I know you know what you meant; but for a newcomer it might not be so obvious. Anyway, I gave you +1 :)
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MarekJun 18 '11 at 16:18

@Marek: actually, I really appreciated your pedantry in this case. It is my mistake for being sloppy. Doing some editing now.
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Willie WongJun 18 '11 at 16:29

Think of a liquid that is flowing in two dimensions on a flat surface. At every point we can draw a vector that points in the direction that the fluid is flowing in at that point. That is a vector in the vector space at that point, the tangent space. There is a tangent space at every point, because at every point the fluid could be flowing in any direction, although an equation of motion might require, for example, continuity from point to point. An example of a vector flow (found through Google images for "fluid flow") is http://www.dstu.univ-montp2.fr/geofracnet/Images/connollyflownet2.jpg. Graphically, of course, the tangent vectors to the flow can only be displayed at a finite number of points, but from the point of view of a differential equation there's a flow vector at every point. All the tangent spaces are exactly the same as all the others.

The "vectors" used to describe the points are vectors in the vector space at one particular point, the center of the coordinate system, with the vector describing the location of the other point relative to the center of the coordinate system as distances in two directions that aren't parallel. The four-dimensional case is no different provided we concern ourselves only with vector spaces, but the idea of a Lorentzian distance is of course different from the idea of a Euclidean distance. In coordinate terms, we could talk of two functions, $V_1(x,y)$ and $V_2(x,y)$, where $x$ and $y$ are components of the position of a point, and $V_1(x,y)$ and $V_2(x,y)$ are components of a vector at that point.

A vector in 2 or in 4 dimensions can of course represent something other than a velocity field. In due course you will get straight what can be represented by vectors and what requires tensors.

Now imagine a fluid flowing on the surface of a sphere. Again there is a tangent space at every point on the surface of the sphere, and all the tangent spaces are exactly the same as the others, but now it's harder to describe explicitly the way in which the many tangent spaces are the same as each other. Saying which point is which by using vectors in a tangent space at a specific point is also more problematic.